$\overline{AB} = 3\sqrt{10}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $3\sqrt{10}$ $?$ $ \sin( \angle ABC ) = \frac{ \sqrt{10}}{10}, \cos( \angle ABC ) = \frac{3\sqrt{10} }{10}, \tan( \angle ABC ) = \dfrac{1}{3}$
Answer: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is adjacent to $\angle ABC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle ABC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{3\sqrt{10}} $ $ \overline{BC}=3\sqrt{10} \cdot \cos( \angle ABC ) = 3\sqrt{10} \cdot \frac{3\sqrt{10} }{10} = 9$